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Shtirlitz [24]
3 years ago
14

Does toothpaste have matter

Physics
1 answer:
miv72 [106K]3 years ago
7 0
Yes. Almost allthings have matter

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WINSTONCH [101]
Brushes, battery terminals, commutator, armature, magnets
7 0
2 years ago
A force of 20 N is exerted on a box with a mass of 15 kg. if friction exerts a force of 4 N on the box, at what rate does the bo
MakcuM [25]

Answer:

1.06 metres per second squared

Explanation:

since friction acts against foward force

20 N - 4 N = 16 N

use Newtons 2nd law F=ma Solve for a:

a= F÷m

= 16 ÷ 15

= 1.06 metres per second squared

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3 years ago
A concave mirror has a radius of curvature of 1.6m. Find the focal length
Yuki888 [10]
Given,

Radius of curvature of concave mirror = 1.6m

We know that ,

Focal length = radius/2

Hence ,

Focal length of concave mirror = radius of concave mirror /2

=> F = 1.6/2

=> F = 0.8m

Hence the focal length of concave mirror is 0.8 m
5 0
2 years ago
Read 2 more answers
Which embedded molecule in the cell membrane moves large molecules into and out of the cell?
Nezavi [6.7K]


The correct  answer is A.

The cell membrane consists of a phospholipid bilayer with embedded proteins. Sometimes molecules are just too big to easily flow across the plasma membrane or dissolve in the water so that they can be filtered through the cell membrane. In these  cases , the cells must put out a little energy to help get molecules in and out of the cell.

The proteins embedded in the plasma membrane form channels through which other molecules can pass. Some proteins act as carriers, that is they are 'paid" in energy to let a molecule attach to itself and  then transport that molecule inside the cell. This is called active transport.



5 0
3 years ago
At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107
dangina [55]

Answer:

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Explanation:

The bulk modulus is represented by the following differential equation:

K = - V\cdot \frac{dP}{dV}

Where:

K - Bulk module, measured in pascals.

V - Sample volume, measured in cubic meters.

P - Local pressure, measured in pascals.

Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:

-\frac{K \,dV}{V} = dP

This resultant expression is solved by definite integration and algebraic handling:

-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP

-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}

\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}

\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }

The final volume is predicted by:

V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }

If V_{o} = 1\,m^{3}, P_{o} - P_{f} = -10132500\,Pa and K = 2.3\times 10^{9}\,Pa, then:

V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }

V_{f} \approx 0.996\,m^{3}

Change in volume due to increasure on pressure is:

\Delta V = V_{o} - V_{f}

\Delta V = 1\,m^{3} - 0.996\,m^{3}

\Delta V = 0.004\,m^{3}

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

8 0
3 years ago
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